Integrand size = 30, antiderivative size = 150 \[ \int \frac {x^{-1+3 n}}{\sqrt {a+b x^n} \sqrt {c+d x^n}} \, dx=-\frac {3 (b c+a d) \sqrt {a+b x^n} \sqrt {c+d x^n}}{4 b^2 d^2 n}+\frac {x^n \sqrt {a+b x^n} \sqrt {c+d x^n}}{2 b d n}-\frac {\left (4 a b c d-3 (b c+a d)^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x^n}}{\sqrt {b} \sqrt {c+d x^n}}\right )}{4 b^{5/2} d^{5/2} n} \]
-1/4*(4*a*b*c*d-3*(a*d+b*c)^2)*arctanh(d^(1/2)*(a+b*x^n)^(1/2)/b^(1/2)/(c+ d*x^n)^(1/2))/b^(5/2)/d^(5/2)/n-3/4*(a*d+b*c)*(a+b*x^n)^(1/2)*(c+d*x^n)^(1 /2)/b^2/d^2/n+1/2*x^n*(a+b*x^n)^(1/2)*(c+d*x^n)^(1/2)/b/d/n
Time = 0.43 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.05 \[ \int \frac {x^{-1+3 n}}{\sqrt {a+b x^n} \sqrt {c+d x^n}} \, dx=\frac {b \sqrt {d} \sqrt {a+b x^n} \left (c+d x^n\right ) \left (-3 b c-3 a d+2 b d x^n\right )+\sqrt {b c-a d} \left (3 b^2 c^2+2 a b c d+3 a^2 d^2\right ) \sqrt {\frac {b \left (c+d x^n\right )}{b c-a d}} \text {arcsinh}\left (\frac {\sqrt {d} \sqrt {a+b x^n}}{\sqrt {b c-a d}}\right )}{4 b^3 d^{5/2} n \sqrt {c+d x^n}} \]
(b*Sqrt[d]*Sqrt[a + b*x^n]*(c + d*x^n)*(-3*b*c - 3*a*d + 2*b*d*x^n) + Sqrt [b*c - a*d]*(3*b^2*c^2 + 2*a*b*c*d + 3*a^2*d^2)*Sqrt[(b*(c + d*x^n))/(b*c - a*d)]*ArcSinh[(Sqrt[d]*Sqrt[a + b*x^n])/Sqrt[b*c - a*d]])/(4*b^3*d^(5/2) *n*Sqrt[c + d*x^n])
Time = 0.28 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.01, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {948, 101, 27, 90, 66, 221}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {x^{3 n-1}}{\sqrt {a+b x^n} \sqrt {c+d x^n}} \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {\int \frac {x^{2 n}}{\sqrt {b x^n+a} \sqrt {d x^n+c}}dx^n}{n}\) |
\(\Big \downarrow \) 101 |
\(\displaystyle \frac {\frac {\int -\frac {3 (b c+a d) x^n+2 a c}{2 \sqrt {b x^n+a} \sqrt {d x^n+c}}dx^n}{2 b d}+\frac {x^n \sqrt {a+b x^n} \sqrt {c+d x^n}}{2 b d}}{n}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {x^n \sqrt {a+b x^n} \sqrt {c+d x^n}}{2 b d}-\frac {\int \frac {3 (b c+a d) x^n+2 a c}{\sqrt {b x^n+a} \sqrt {d x^n+c}}dx^n}{4 b d}}{n}\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {\frac {x^n \sqrt {a+b x^n} \sqrt {c+d x^n}}{2 b d}-\frac {\frac {\left (4 a b c d-3 (a d+b c)^2\right ) \int \frac {1}{\sqrt {b x^n+a} \sqrt {d x^n+c}}dx^n}{2 b d}+\frac {3 (a d+b c) \sqrt {a+b x^n} \sqrt {c+d x^n}}{b d}}{4 b d}}{n}\) |
\(\Big \downarrow \) 66 |
\(\displaystyle \frac {\frac {x^n \sqrt {a+b x^n} \sqrt {c+d x^n}}{2 b d}-\frac {\frac {\left (4 a b c d-3 (a d+b c)^2\right ) \int \frac {1}{b-d x^{2 n}}d\frac {\sqrt {b x^n+a}}{\sqrt {d x^n+c}}}{b d}+\frac {3 (a d+b c) \sqrt {a+b x^n} \sqrt {c+d x^n}}{b d}}{4 b d}}{n}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\frac {x^n \sqrt {a+b x^n} \sqrt {c+d x^n}}{2 b d}-\frac {\frac {\left (4 a b c d-3 (a d+b c)^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x^n}}{\sqrt {b} \sqrt {c+d x^n}}\right )}{b^{3/2} d^{3/2}}+\frac {3 (a d+b c) \sqrt {a+b x^n} \sqrt {c+d x^n}}{b d}}{4 b d}}{n}\) |
((x^n*Sqrt[a + b*x^n]*Sqrt[c + d*x^n])/(2*b*d) - ((3*(b*c + a*d)*Sqrt[a + b*x^n]*Sqrt[c + d*x^n])/(b*d) + ((4*a*b*c*d - 3*(b*c + a*d)^2)*ArcTanh[(Sq rt[d]*Sqrt[a + b*x^n])/(Sqrt[b]*Sqrt[c + d*x^n])])/(b^(3/2)*d^(3/2)))/(4*b *d))/n
3.11.75.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 2 Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre eQ[{a, b, c, d}, x] && !GtQ[c - a*(d/b), 0]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( p_), x_] :> Simp[b*(a + b*x)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 3))), x] + Simp[1/(d*f*(n + p + 3)) Int[(c + d*x)^n*(e + f*x)^p*Simp [a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f *(n + p + 4) - b*(d*e*(n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. ), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^ p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ [b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
\[\int \frac {x^{-1+3 n}}{\sqrt {a +b \,x^{n}}\, \sqrt {c +d \,x^{n}}}d x\]
Time = 0.30 (sec) , antiderivative size = 361, normalized size of antiderivative = 2.41 \[ \int \frac {x^{-1+3 n}}{\sqrt {a+b x^n} \sqrt {c+d x^n}} \, dx=\left [\frac {{\left (3 \, b^{2} c^{2} + 2 \, a b c d + 3 \, a^{2} d^{2}\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2 \, n} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, \sqrt {b d} b d x^{n} + {\left (b c + a d\right )} \sqrt {b d}\right )} \sqrt {b x^{n} + a} \sqrt {d x^{n} + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x^{n}\right ) + 4 \, {\left (2 \, b^{2} d^{2} x^{n} - 3 \, b^{2} c d - 3 \, a b d^{2}\right )} \sqrt {b x^{n} + a} \sqrt {d x^{n} + c}}{16 \, b^{3} d^{3} n}, -\frac {{\left (3 \, b^{2} c^{2} + 2 \, a b c d + 3 \, a^{2} d^{2}\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, \sqrt {-b d} b d x^{n} + {\left (b c + a d\right )} \sqrt {-b d}\right )} \sqrt {b x^{n} + a} \sqrt {d x^{n} + c}}{2 \, {\left (b^{2} d^{2} x^{2 \, n} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x^{n}\right )}}\right ) - 2 \, {\left (2 \, b^{2} d^{2} x^{n} - 3 \, b^{2} c d - 3 \, a b d^{2}\right )} \sqrt {b x^{n} + a} \sqrt {d x^{n} + c}}{8 \, b^{3} d^{3} n}\right ] \]
[1/16*((3*b^2*c^2 + 2*a*b*c*d + 3*a^2*d^2)*sqrt(b*d)*log(8*b^2*d^2*x^(2*n) + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 4*(2*sqrt(b*d)*b*d*x^n + (b*c + a*d)*sq rt(b*d))*sqrt(b*x^n + a)*sqrt(d*x^n + c) + 8*(b^2*c*d + a*b*d^2)*x^n) + 4* (2*b^2*d^2*x^n - 3*b^2*c*d - 3*a*b*d^2)*sqrt(b*x^n + a)*sqrt(d*x^n + c))/( b^3*d^3*n), -1/8*((3*b^2*c^2 + 2*a*b*c*d + 3*a^2*d^2)*sqrt(-b*d)*arctan(1/ 2*(2*sqrt(-b*d)*b*d*x^n + (b*c + a*d)*sqrt(-b*d))*sqrt(b*x^n + a)*sqrt(d*x ^n + c)/(b^2*d^2*x^(2*n) + a*b*c*d + (b^2*c*d + a*b*d^2)*x^n)) - 2*(2*b^2* d^2*x^n - 3*b^2*c*d - 3*a*b*d^2)*sqrt(b*x^n + a)*sqrt(d*x^n + c))/(b^3*d^3 *n)]
\[ \int \frac {x^{-1+3 n}}{\sqrt {a+b x^n} \sqrt {c+d x^n}} \, dx=\int \frac {x^{3 n - 1}}{\sqrt {a + b x^{n}} \sqrt {c + d x^{n}}}\, dx \]
\[ \int \frac {x^{-1+3 n}}{\sqrt {a+b x^n} \sqrt {c+d x^n}} \, dx=\int { \frac {x^{3 \, n - 1}}{\sqrt {b x^{n} + a} \sqrt {d x^{n} + c}} \,d x } \]
\[ \int \frac {x^{-1+3 n}}{\sqrt {a+b x^n} \sqrt {c+d x^n}} \, dx=\int { \frac {x^{3 \, n - 1}}{\sqrt {b x^{n} + a} \sqrt {d x^{n} + c}} \,d x } \]
Timed out. \[ \int \frac {x^{-1+3 n}}{\sqrt {a+b x^n} \sqrt {c+d x^n}} \, dx=\int \frac {x^{3\,n-1}}{\sqrt {a+b\,x^n}\,\sqrt {c+d\,x^n}} \,d x \]